Advertisements
Advertisements
प्रश्न
Evaluate the following :
`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
Advertisements
उत्तर
`lim_(x -> 0) (x(6^x - 3^x))/(cos (6x) - cos (4x))`
= `lim_(x -> 0) (x*3^x (2^x - 1))/(-2sin ((6x + 4x)/2) sin((6x - 4x)/2))`
= `1/(-2) lim_(x -> 0) (x*3^x (2^x - 1))/(sin 5x*sinx)`
= `1/(-2) lim_(x -> 0) ((x*3^x (2^x - 1))/x^2)/((sin 5x* sinx)/x^2) ...[("Divide Numerator and"),("Denominator by" x^2),(∵ x -> 0"," ∴ x ≠ 0 ∴ x^2 ≠ 0)]`
= `1/(-2) (lim_(x -> 0) [3^x ((2^x - 1))/x])/(lim_(x -> 0) ((sin5x)/x * sinx/x))`
= `1/(-2) (lim_(x -> 0)(3x) * lim_(x -> 0) ((2^x - 1)/x))/(lim_(x -> 0) (sinx/(5x) xx 5) * lim_(x -> 0) (sinx/x))`
= `1/(-2) xx (3^circ xx log 2)/(1 xx 5 xx 1) ...[(because x -> 0"," 5x -> 0),(lim_(x -> 0) ("a"^x - 1)/x = log "a"),(lim_(x -> 0) sinx/x = 1)]`
= `(-1)/10 log 2`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x ->0) cos x/(pi - x)`
Evaluate the following limit.
`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`
Evaluate the following limit.
`lim_(x → 0) x sec x`
Evaluate the following limit :
`lim_(x -> pi/6) [(2 - "cosec"x)/(cot^2x - 3)]`
Select the correct answer from the given alternatives.
`lim_(x → π/3) ((tan^2x - 3)/(sec^3x - 8))` =
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
Evaluate `lim_(x -> 2) 1/(x - 2) - (2(2x - 3))/(x^3 - 3x^2 + 2x)`
Evaluate `lim_(x -> pi/2) (secx - tanx)`
Evaluate `lim_(x -> 0) (sin(2 + x) - sin(2 - x))/x`
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
`lim_(x -> pi/2) (1 - sin x)/cosx` is equal to ______.
If f(x) = x sinx, then f" `pi/2` is equal to ______.
Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`
Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`
Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`
Evaluate: `lim_(x -> pi/3) (sqrt(1 - cos 6x))/(sqrt(2)(pi/3 - x))`
Evaluate: `lim_(x -> pi/4) (sin x - cosx)/(x - pi/4)`
Evaluate: `lim_(x -> 0) (sin 2x + 3x)/(2x + tan 3x)`
x cos x
`lim_(x -> pi) (1 - sin x/2)/(cos x/2 (cos x/4 - sin x/4))`
Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists
`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.
`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.
`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
`lim_(x -> 0) (sin mx cot x/sqrt(3))` = 2, then m = ______.
`lim_(x -> 3^+) x/([x])` = ______.
